## Proof of Divergence Theorem

Theorem
The divergence theorem states that if
$\mathbf{F}$
is a vector field in a volume
$V$
bounded by a surface
$S$
then
$\int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV = \int \int_S \mathbf{F} \cdot \mathbf{n} dS$

Proof Let
$\mathbf{F}= F_1 \mathbf{i} + F_2 \mathbf{j} + F_3 \mathbf{k}$

\begin{aligned} \int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV &= \int \int \int_V \frac{\partial^2 F_1}{\partial x^2} + \frac{\partial^2 F_2}{\partial y^2} + \frac{\partial^2 F_3}{\partial z^2} \: dx \: dy \: dz \\ &= \int \int_{yz} F_! (x_{FRONT SURFACE},y,z ) -F_! (x_{BACK SURFACE},y,z ) dy \: dz \\ &= \int \int_{xz} F_! (x,y_{RIGHT SURFACE},z ) -F_! (x,y_{LEFT SURFACE},z ) dx \: dz \\ &= \int \int_{xy} F_! (x,y,z_{TOP SURFACE} ) -F_! (x,y,z_{BOTTOM SURFACE} ) dx \: dy \end{aligned}

On the front and back surfaces
$dydz= cos \alpha_{FRONT} dS_{FRONT} = \mathbf{i} \cdot \mathbf{n} dS_{FRONT}, \: dydz=- cos \alpha_{BACK} dS_{FRONT}=-\mathbf{i} \cdot \mathbf{n} dS_{BACk}$
respectively.
On the Left and right surfaces
$dxdz= cos \alpha_{RIGHT} dS_{RIGHT} = \mathbf{j} \cdot \mathbf{n} dS_{LEFT}, \: dydz=- cos \alpha_{LEFT} dS_{LEFT} = -\mathbf{j} \cdot \mathbf{n} dS_{RIGHT}$
respectively.
On the top and bottom surfaces
$dxdy= cos \alpha_{TOP} dS_{TOP} =\mathbf{k} \cdot \mathbf{n} dS_{TOP}, \: dydz=- cos \alpha_{BOTTOM} dS_{BOTTOM} =- \mathbf{k} \cdot \mathbf{n} dS_{BOTTOM}$
respectively.
Hence
$\int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV = \int \int_S \mathbf{F} \cdot \mathbf{n} dS$